Q: \$PQ\$ is a chord of length \$8\$ cm of a circle of radius \$5\$ cm. The tangents \$P\$ and \$Q\$ intersect at a point \$T\$. Find the length \$TP\$.

Joint \$OT\$. Let it meet \$PQ\$ at the point \$R\$. Then \$\Delta TPQ\$ is isosceles and \$TO\$ is the angle bisector of \$\angle PTO\$.[\$\because TP = TQ =\$ Tangents from \$T\$ upon the circle]\$\therefore OT \perp PQ\$\$\therefore OT\$ bisects \$PQ\$.\$PR =RQ = 4\$ cmNow, \$OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3\$ cmNow, \$\angle TPR + \angle RPO = 90^{\circ} (\because TPO = 90^{\circ})\$\$= \angle TPR + \angle PTR (\because TRP = 90^{\circ})\$\$\therefore \angle RPO = \angle PTR\$\$\therefore\$ Right triangle \$TRP\$ is similar to the right triangle \$PRO\$. [By A-A Rule of similar triangles]\$\displaystyle \therefore \frac{TP}{PO} = \frac{RP}{RO} \Rightarrow \frac{TP}{5} = \frac{4}{3}\$\$\displaystyle \Rightarrow TP = \frac{20}{3}\$ cm.

Q: If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Given:\$ABCD\$ is a trapezium where non-parallel sides \$AD\$ and \$BC\$ are equal. Construction:\$DM\$ and \$CN\$ are perpendicular drawn on \$AB\$ from \$D\$ and \$C\$, respectively. To prove:\$ABCD\$ is cyclic trapezium.Proof: In \$\triangle DAM\$ and \$\triangle CBN\$,\$AD = BC\$ ...[Given]\$∠AMD = ∠BNC\$ ...[Right angles]\$DM = CN\$ ...[Distance between the parallel lines]Therefore, \$\triangle DAM ≅ \triangle CBN\$ by \$RHS\$ congruence condition. Now, \$∠A = ∠B\$ ...[by CPCT]Also, \$∠B + ∠C = 180°\$ ....[Sum of the co-interior angles]\$⇒ ∠A + ∠C = 180°\$. Thus, \$ABCD\$ is a cyclic quadrilateral as the sum of the pair of opposite angles is \$180°\$.

Q: A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Consider the figure.Given,\$AB\$ is equal to the radius of the circle. In \$\triangle OAB\$,\$OA = OB = AB =\$ radius of the circle. Thus, \$\triangle OAB\$ is an equilateral triangle.and \$∠AOC = 60°\$.Also, \$∠ACB = \dfrac {1}{2}∠AOB = \dfrac {1}{2} × 60° = 30°\$.Since, \$ACBD\$ is a cyclic quadrilateral,\$∠ACB + ∠ADB = 180°\$ ....[Opposite angles of cyclic quadrilateral are supplementary]\$⇒ ∠ADB = 180° - 30° = 150°\$. Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are \$150°\$ and \$30°\$, respectively.

Q: Prove that a cyclic parallelogram is a rectangle.

Given,\$ABCD\$ is a cyclic parallelogram.To prove,\$ABCD\$ is a rectangle.Proof:\$∠1 + ∠2 = 180°\$ ...Opposite angles of a cyclic parallelogramAlso, Opposite angles of a cyclic parallelogram are equal.Thus,\$∠1 = ∠2\$\$⇒ ∠1 + ∠1 = 180°\$\$⇒ ∠1 = 90°\$One of the interior angle of the parallelogram is right angled. Thus, \$ABCD\$ is a rectangle.

Q: In given fig. O is centre of the circle if \$\angle AOC={ 130 }^{ o }\$ then find \$\angle ABC\$.

\$\angle ADC = \dfrac{1}{2} \angle AOC\$(Angle subtended by a chord on mayor area is half of that subtended on center)\$\therefore \angle ADC = \dfrac{1}{2} \times 130 = 65^{\circ}\$Now, ADCB is cyclic quadrilateral \$\Rightarrow \$ Sum of opposite Angles is \$180^{\circ}\$\$\therefore \angle ABC + \angle ADC = 180^{\circ}\$\$\angle ABC + 65= 180^{\circ}\$\$\angle ABC = 115^{\circ}\$

Q: A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and \$\displaystyle \angle ADC=130^{\circ}\$ Find \$\displaystyle \angle ABC \$ (in degrees)

\$\Rightarrow\$ Quadrilateral \$ABCD\$ inscribe in a circle.\$\therefore\$ Quadrilateral \$ABCD\$ is a cyclic quadrilateral.\$\Rightarrow\$ \$\angle ADC+\angle ABC=180^o\$ [ Sum of opposite angles of cyclic quadrilateral is supplementary ]\$\Rightarrow\$ \$130^o+\angle ABC=180^o\$\$\Rightarrow\$ \$\angle ABC=180^o-130^o\$\$\Rightarrow\$ \$\angle ABC=50^o\$

Q: \$ABCD\$ is a cyclic quadrilateral such that \$ A={ 90 }^{ o }, B={ 70 }^{ o }, C={ 95 }^{ o } \$ and \$D={ 105 }^{ o }.\$The statement is:

Given, \$ABCD\$ is a cyclic quadrilateral where \$ \angle A={ 90 }^{ o } , \angle B={ 70 }^{ o }, \angle C={ 95 }^{ o }\$ & \$\angle D={ 105 }^{ o }.\$Since \$ABCD\$ is a cyclic quadrilateral,then \$\angle A+\angle C ={ 180 }^{ o }\$ and \$ \angle B+\angle D={ 180 }^{ o }\$......\$(i)\$(The sum of the opposite angles of a cyclic quadrilateral is \${ 180 }^{ o })\$But here, \$ \angle A+\angle C={ 90 }^{ o }+95^{ o }=185^{ o }\$ and \$\angle B+\angle D={ 70 }^{ o }+{ 10 }5^{ o }=175^{ o }.\$This is a contradiction to \$(i)\$.That is, the statement is false.Hence, option \$B\$ is correct.

Question 1

$PQ$ is a chord of length $8$ cm of a circle of radius $5$ cm. The tangents $P$ and $Q$ intersect at a point $T$. Find the length $TP$.

Accepted Answer

Joint $OT$. Let it meet $PQ$ at the point $R$. Then $\Delta TPQ$ is isosceles and $TO$ is the angle bisector of $\angle PTO$.[$\because TP = TQ =$ Tangents from $T$ upon the circle]$\therefore OT \perp PQ$$\therefore OT$ bisects $PQ$.$PR =RQ = 4$ cmNow, $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3$ cmNow, $\angle TPR + \angle RPO = 90^{\circ} (\because TPO = 90^{\circ})$$= \angle TPR + \angle PTR    (\because TRP = 90^{\circ})$$\therefore \angle RPO = \angle PTR$$\therefore$ Right triangle $TRP$ is similar to the right triangle $PRO$. [By A-A Rule of similar triangles]$\displaystyle \therefore \frac{TP}{PO} = \frac{RP}{RO} \Rightarrow \frac{TP}{5} = \frac{4}{3}$$\displaystyle \Rightarrow TP = \frac{20}{3}$ cm.

Question 2

Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle .

Accepted Answer

Given :An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle.To prove : $\angle POQ =2\angle PAQ$To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB.Construction :Join the line AO extended to B.Proof :$\angle BOQ = \angle OAQ+\angle AQO$       .....(1)Also, in $\triangle$ OAQ,$OA=OQ$                  [Radii of a circle]Therefore,$\angle OAQ = \angle OQA$      [Angles opposite to equal sides are equal]$\angle BOQ = 2\angle OAQ$              .......(2)Similarly, $BOP=2\angle OAP$             ........(3)Adding 2 & 3, we get,$\angle BOP +\angle BOQ = 2(\angle OAP+\angle OAQ)$$\angle POQ = 2\angle PAQ$               .......(4)For the case 3, where PQ is the major arc, equation 4 is replaced byReflex angle, $\angle POQ=2\angle PAQ$

Question 3

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Accepted Answer

Given:$ABCD$ is a trapezium where non-parallel sides $AD$ and $BC$ are equal.&#10;Construction:$DM$ and $CN$ are perpendicular drawn on $AB$ from $D$ and $C$, respectively.&#10;To prove:$ABCD$ is cyclic trapezium.Proof:&#10;In $\triangle DAM$ and $\triangle CBN$,$AD = BC$       ...[Given]$&#8736;AMD = &#8736;BNC$      ...[Right angles]$DM = CN$     ...[Distance between the parallel lines]Therefore, $\triangle DAM &#8773; \triangle CBN$  by $RHS$ congruence condition.&#10;Now, $&#8736;A = &#8736;B$      ...[by CPCT]Also, $&#8736;B + &#8736;C = 180&#176;$    ....[Sum of the co-interior angles]$&#8658; &#8736;A + &#8736;C = 180&#176;$.&#10;Thus, $ABCD$ is a cyclic quadrilateral as the sum of the pair of opposite angles is $180&#176;$.

Question 4

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Accepted Answer

Consider the figure.Given,$AB$ is equal to the radius of the circle.&#10;In $\triangle OAB$,$OA = OB = AB =$ radius of the circle.&#10;Thus, $\triangle OAB$ is an equilateral triangle.and $&#8736;AOC = 60&#176;$.Also, $&#8736;ACB = \dfrac {1}{2}&#8736;AOB = \dfrac {1}{2} &#215; 60&#176; = 30&#176;$.Since, $ACBD$ is a cyclic quadrilateral,$&#8736;ACB + &#8736;ADB = 180&#176;$ ....[Opposite angles of cyclic quadrilateral are supplementary]$&#8658; &#8736;ADB = 180&#176; - 30&#176; = 150&#176;$.&#10;Thus, angle subtend by the chord at a point on the minor arc and also at a point on the major arc are $150&#176;$ and $30&#176;$, respectively.&#10;

Question 5

The perpendicular from the centre of a circle to a chord bisects the chord.

Accepted Answer

To prove that the perpendicular from the centre to a chord bisect the chord.Consider a circle with centre at $O$ and $AB$ is a chord such that $OX$ perpendicular to $AB$To prove that   $AX = BX$In $ \Delta OAX $ and $ \Delta OBX $$ \angle OXA = \angle OXB $  [both are 90 ]$  OA = OB $  (Both  are radius of circle )$  OX = OX $  (common side )$&#916;OAX\cong &#916;OBX$$AX = BX$  (by property of congruent triangles )hence proved.

Question 6

Prove that a cyclic parallelogram is a rectangle.

Accepted Answer

Given,$ABCD$ is a cyclic parallelogram.To prove,$ABCD$ is a rectangle.Proof:$&#8736;1 + &#8736;2 = 180&#176;$      ...Opposite angles of a cyclic parallelogramAlso, Opposite angles of a cyclic parallelogram are equal.Thus,$&#8736;1 = &#8736;2$$&#8658; &#8736;1 + &#8736;1 = 180&#176;$$&#8658; &#8736;1 = 90&#176;$One of the interior angle of the parallelogram is right angled. Thus, $ABCD$ is a rectangle.&#10;

Question 7

In given fig. O is centre of the circle if $\angle AOC={ 130 }^{ o }$ then find $\angle ABC$.

Accepted Answer

$\angle ADC = \dfrac{1}{2} \angle AOC$(Angle subtended by a chord on mayor area is half of that subtended on center)$\therefore \angle ADC = \dfrac{1}{2} \times 130 = 65^{\circ}$Now, ADCB is cyclic quadrilateral $\Rightarrow $ Sum of opposite Angles is $180^{\circ}$$\therefore \angle ABC + \angle ADC = 180^{\circ}$$\angle ABC + 65= 180^{\circ}$$\angle ABC = 115^{\circ}$

Question 8

In Fig 10.4 if $\displaystyle \angle ABC = 20^{\circ}$ then $\displaystyle \angle AOC $ is equal to

Accepted Answer

We know that the angle at the centre of the circle is twice the angle at the circumference subtended by the same arc.Thus,   $\angle AOC=2 \times \angle ABC$$\implies \angle AOC=2\times 20^0$$\implies \angle AOC=40^0$

Question 9

A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and $\displaystyle \angle ADC=130^{\circ}$ Find $\displaystyle \angle ABC $ (in degrees)

Accepted Answer

$\Rightarrow$  Quadrilateral $ABCD$ inscribe in a circle.$\therefore$  Quadrilateral $ABCD$ is a cyclic quadrilateral.$\Rightarrow$  $\angle ADC+\angle ABC=180^o$           [ Sum of opposite angles of cyclic quadrilateral is supplementary ]$\Rightarrow$  $130^o+\angle ABC=180^o$$\Rightarrow$  $\angle ABC=180^o-130^o$$\Rightarrow$  $\angle ABC=50^o$

Question 10

$ABCD$ is a cyclic quadrilateral such that $ A={ 90 }^{ o }, B={ 70 }^{ o }, C={ 95 }^{ o } $ and $D={ 105 }^{ o }.$The statement is:

Accepted Answer

Given, $ABCD$ is a cyclic quadrilateral where $ \angle A={ 90 }^{ o } , \angle B={ 70 }^{ o }, \angle C={ 95 }^{ o }$ &  $\angle D={ 105 }^{ o }.$Since $ABCD$ is a cyclic quadrilateral,then $\angle A+\angle C ={ 180 }^{ o }$ and $ \angle B+\angle D={ 180 }^{ o }$......$(i)$(The sum of the opposite angles of a cyclic quadrilateral is ${ 180 }^{ o })$But here, $ \angle A+\angle C={ 90 }^{ o }+95^{ o }=185^{ o }$ and $\angle B+\angle D={ 70 }^{ o }+{ 10 }5^{ o }=175^{ o }.$This is a contradiction to $(i)$.That is, the statement is false.Hence, option $B$ is correct.